Probability

Since

$\displaystyle \lfloor \sqrt{P(x)}\rfloor=\sqrt{P(\lfloor x \rfloor)} \in \mathbb{N} \hspace{10pt} (\star)$ ,

than the only $x \in A=[5,15]$ satisfying $\sqrt{P(x)} \in \mathbb{N}$ are $x_1=5$ , $x_2=6$ , $x_3=13$ . Let's explain the meaning of $(\star)$ with an example: we see that $\sqrt{P(5)}=1$ . We also see that $\sqrt{P(5.3)}=1.78605...$ , so $\lfloor \sqrt{P(5.3)}\rfloor=1$ , satisfying $(\star)$ . Our goal is to find all the $x_i+\epsilon$ , $\epsilon <1$ , $i=\{1,2,3\}$ such that $(\star)$ holds. For example, to find all the $5+\epsilon$ , we impose that

$\displaystyle \sqrt{x^2-3x-9}=2 \hspace{5pt} \Longrightarrow \hspace{5pt} x=\frac{1}{2}(3+\sqrt{61}) \approx5.4051...$

It means that, for every $\displaystyle x \in L_1= \bigg[5,\frac{1}{2}(3+\sqrt{61})\bigg)$ , $(\star)$ holds. Let's apply the same procedure to $x_2$ and $x_3$ ; we get

$\displaystyle L_2=\bigg[6,\frac{1}{2}(3+\sqrt{109})\bigg)$

$\displaystyle L_3=\bigg[13,\frac{1}{2}(3+3\sqrt{69})\bigg)$

If we call $L=L_1 \cup L_2 \cup L_3$ , than

$\displaystyle \mathbb{P}(L)=\frac{Lenght(L)}{Lenght(A)}=\frac{(\frac{1}{2}(3+\sqrt{61})-5)+(\frac{1}{2}(3+\sqrt{109})-6)+(\frac{1}{2}(3+3\sqrt{69})-13)}{10}=\frac{\sqrt{109}+\sqrt{61}+\sqrt{621}-39}{20}$

Eventually,

$a=109$ , $b=61$ , $c=621$ , $d=39$ , $e=20$ , $\hspace{10pt} a+b+c+d+e=\boxed{850}$

Looking at the left side of the equation, we know that the value of the equation is an integer, which means that $\sqrt{P \left( \left \lfloor{x}\right \rfloor \right) }$ is an integer. $\left \lfloor{x}\right \rfloor$ must assume any value of {5, 6, 7, ..., 12, 13, 14}, because $5 \leq x \leq 15$ , so we evaluate $\sqrt{P(k)}$ for $k$ in of {5, 6, 7, ..., 12, 13, 14} and we see that only for $k = 5$ , $k = 6$ and $k = 13$ that value is integer.

That means that $\left \lfloor{ \sqrt{P(x)} }\right \rfloor$ must be $1$ , $3$ or $11$ , so, $\sqrt{P(x)}$ will be some value in $[1,2]$ , $[3,4]$ or $[11,12]$ and $P(x)$ will be some value in $[1,4]$ , $[9,16]$ or $[121,144]$ . Solving the three inequalities we obtain the answer.